Conditional expectation of three exponential distributed r.v.

[Ego7894612]

I've been struggling with this problem for more than 4 days now:

Let A, B and C be exponential distributed random variables with parameters lambda_A, lambda_B and lambda_C, respectively.

Calculate E [ B | A < B < C ] in terms of the lambda's.

I always seem get an integral which is impossible to calculate...

Who has a suggestion how to solve this problem?

 

[bpet]

I always seem get an integral which is impossible to calculate...

That's how I'd try to do it too, can you show us what you got and where you were stuck?

 

[Ego7894612]

First a remark: we may assume that A, B and C are independent random variables (this may help a lot).

<br /> E [ B | A &lt; B &lt; C ] = \int_{0}^\infty E [ B \, | \, A &lt; B &lt; C, A = a ] \; f_{A} ( a ) \; da <br />
<br /> = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, A &lt; B &lt; C, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc.<br />
(Along with all that follows, this integral is only to be considered over R_+^2 and for c>a.)
<br /> = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a &lt; B &lt; c, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc<br />
(by independence)
<br /> = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a &lt; B &lt; c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc<br />

We want to compute E [ B \, | \, a &lt; B &lt; c ].

First note that
E [ B \, | \, a &lt; B &lt; c ] = \frac{ \int_a^c t f_{B} (t) dt }{P( B \in [a,c])}.

We have\int_a^c t f_{B} (t) dt = ... = (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} )
and
P(B \in [a,c]) = F_B (c) - F_B (a) = ... = e^{- \lambda_B a} - e^{- \lambda_B c}.

Hence
E [ B \, | \, a &lt; B &lt; c ] = \frac{ (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} ) }{e^{- \lambda_B a} - e^{- \lambda_B c}},
E [ B \, | \, a &lt; B &lt; c ] = \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B}.

So
<br /> E [ B | A &lt; B &lt; C ] = \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B} \right) \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc<br />
<br /> = \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} \right) \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc + <br />
<br /> \int_{0}^\infty \int_{0}^\infty \frac{1}{\lambda_B} \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc<br />Here's my problem: I can't compute the first of these integrals... And even then I'm not sure if what I'm doing here is correct... Is my calculation right, or is it wrong!? Is there perhaps a better way to calculate this? Etc.

 

[bpet]

<br /> E [ B | A &lt; B &lt; C ] = \int_{0}^\infty E [ B \, | \, A &lt; B &lt; C, A = a ] \; f_{A} ( a ) \; da <br />

I'm not sure about this line - the marginals of the conditional distribution aren't necessarily exponential. Another approach is to apply directly the definition of conditional probabilities, so that

E[B|A&lt;B&lt;C] = E[B.I(A&lt;B&lt;C)]/P[A&lt;B&lt;C]

which is a ratio of integrals that should both be tractable.

Hope this helps.

 

[Ego7894612]

I'm not sure about this line - the marginals of the conditional distribution aren't necessarily exponential. Another approach is to apply directly the definition of conditional probabilities, so that

E[B|A&lt;B&lt;C] = E[B.I(A&lt;B&lt;C)]/P[A&lt;B&lt;C]

which is a ratio of integrals that should both be tractable.

Hope this helps.

Isn't this just the "law of total expectation"?

I don't really understand your point I'm afraid.

EDIT: Yes, we can also go by the second approach.

Then how to calculate E[B.I(A&lt;B&lt;C)] exactly ? (I'm not sure how to start with this integral.)

 

[Ego7894612]

Isn't this just the "law of total expectation"?

I don't really understand your point I'm afraid.

EDIT: Yes, we can also go by the second approach.

Then how to calculate E[B.I(A&lt;B&lt;C)] exactly ? (I'm not sure how to start with this integral.)

I now understand what you mean by "I'm not sure about this line (...)": you're completely right, it's wrong!

 

[bpet]

Then how to calculate E[B.I(A&lt;B&lt;C)] exactly ? (I'm not sure how to start with this integral.)

Ok, this can be simplified to

\int_R b.f_A(a).f_B(b).f_C(c)da.db.dc

where R is the region a<b<c.